Area between curves intersecting more than twice — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Finding all intersection points of two functions, partitioning the integration interval by intersections, identifying upper/lower curves per interval, and calculating total bounded area between two curves with three or more intersections.

You should already know: How to find roots of functions and solve for intersections of two curves. How to evaluate definite integrals via the Fundamental Theorem of Calculus. How to compare function values on a closed interval.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Area between curves intersecting more than twice?

When two curves cross each other more than two times, the upper (top) and lower (bottom) curves swap positions between consecutive intersection points. Unlike area between two curves that intersect only twice, this topic requires splitting the entire interval between the leftmost and rightmost intersection into multiple smaller subintervals, where one curve stays consistently above the other. This topic is part of Unit 8 (Applications of Integration) in the AP Calculus AB CED, which accounts for 10-15% of the total exam score. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQs typically ask to identify the correct integral set-up for total area, while FRQs ask for full numerical calculation, often combined with other application topics. The core extra skill required here is systematically tracking which function is on top in each interval, which is the source of most student errors on this topic.

2. Finding and sorting all intersection points

The non-negotiable first step for any area problem with multiple intersections is finding every solution to $f(x)=g(x)$, then sorting them from left to right (smallest $x$ to largest $x$). Missing even one intersection point leads to incorrect interval splitting, wrong upper/lower assignments, and ultimately an incorrect total area. To find intersections, always rearrange the equation to $f(x)-g(x)=0$ first, then factor the resulting expression (the most common case on AP AB is polynomials) to find all roots. Never divide both sides by a variable term like $x$ before factoring, as this eliminates the $x=0$ root that is a common intersection point. Once all roots are found, sort them to get the boundaries of each subinterval.

Worked Example

Find all sorted $x$-coordinates of intersections between $f(x)=x^3-4x$ and $g(x)=x$.

1. Set the functions equal and rearrange to equal zero: $x^3-4x=x⟹x^3-5x=0$
2. Factor the resulting expression: $x(x^2-5)=0$
3. Solve for all roots: $x=0$, $x^2=5⟹x=-\sqrt{5}$ or $x=\sqrt{5}$
4. Sort roots from left to right: Ordered intersections: $x=-\sqrt{5}\approx -2.236$, $x=0$, $x=\sqrt{5}\approx 2.236$

Exam tip: Always factor out common variable terms first before solving for roots to catch the $x=0$ intersection that most students miss when dividing early.

3. Identifying upper and lower curves per subinterval

After sorting intersection points $x_0<x_1<x_2<...<x_n$, the full interval between the leftmost ($x_0$) and rightmost ($x_n$) intersection is split into $n$ subintervals: $[x_0,x_1],[x_1,x_2],...,[x_{n-1},x_n]$. Between any two consecutive intersections, neither curve crosses the other, so one curve is strictly above the other for the entire subinterval. To confirm which is upper and which is lower, pick any test $x$-value inside the subinterval, then calculate $f(x)-g(x)$. If the difference is positive, $f(x)$ is upper; if negative, $g(x)$ is upper. While polynomials usually alternate upper/lower after each intersection, this is not guaranteed for non-polynomials or higher-degree functions with double roots, so testing is always required.

Worked Example

Identify upper and lower curves for each subinterval between the intersections of $f(x)=x^3-4x$ and $g(x)=x$, which have sorted intersections at $-\sqrt{5},0,\sqrt{5}$.

1. First subinterval $[-\sqrt{5},0]$: Pick test $x=-1$: $f(-1)-g(-1)=((-1{)}^3-4(-1))-(-1)=(-1+4)+1=4>0$ So $f(x)$ is upper, $g(x)$ is lower on this interval.
2. Second subinterval $[0,\sqrt{5}]$: Pick test $x=1$: $f(1)-g(1)=(1-4(1))-1=(-3)-1=-4<0$ So $g(x)$ is upper, $f(x)$ is lower on this interval.

Exam tip: If you have a graphing calculator, plot the curves to confirm your test result, but always show the test step on FRQs to earn full credit for your set-up.

4. Setting up and evaluating total area

Total area between two curves is always positive, because it measures the physical space between the curves, not net signed area. To get a positive area for each subinterval, we always integrate (upper function minus lower function) over each subinterval, then add the results together. The general formula for total area is: $$\text{Total Area}=\sum _{i=1}^n{\int }_{x_{i-1}}^{x_i}\left(\text{upper}(x)-\text{lower}(x)\right)dx={\int }_{x_0}^{x_n}|f(x)-g(x)|dx$$ The absolute value in the second formula guarantees a positive result, and splitting the interval at intersections lets us remove the absolute value by ordering the difference correctly, making integration straightforward. For AP Calculus AB, after setting up the sum of integrals, you just apply the power rule and Fundamental Theorem of Calculus to evaluate each integral, then add the results.

Worked Example

Calculate the total bounded area between $f(x)=x^3-4x$ and $g(x)=x$, using the intersections and upper/lower assignments from the previous examples.

1. Set up the sum of integrals based on our earlier results: $$A={\int }_{-\sqrt{5}}^0\left[(x^3-4x)-x\right]dx+{\int }_0^{\sqrt{5}}\left[x-(x^3-4x)\right]dx$$ Simplify integrands: $$A={\int }_{-\sqrt{5}}^0(x^3-5x)dx+{\int }_0^{\sqrt{5}}(-x^3+5x)dx$$
2. Find antiderivatives: First integral antiderivative: $\frac{x^4}{4}-\frac{5x^2}{2}$ Second integral antiderivative: $-\frac{x^4}{4}+\frac{5x^2}{2}$
3. Evaluate the first definite integral: $\left(0-0\right)-\left(\frac{25}{4}-\frac{5(5)}{2}\right)=-\left(\frac{25}{4}-\frac{25}{2}\right)=\frac{25}{4}$
4. Evaluate the second definite integral: $\left(-\frac{25}{4}+\frac{5(5)}{2}\right)-(0+0)=\frac{25}{4}$
5. Add the results to get total area: $A=\frac{25}{4}+\frac{25}{4}=\frac{25}{2}=12.5$

Exam tip: On FRQs, you earn full credit for a correct integral set-up even before evaluating, so prioritize getting the sum right over rushing to calculate the final number.

Common Pitfalls (and how to avoid them)

• Wrong move: Dividing both sides of $f(x)=g(x)$ by $x$ before factoring, leading to missing the $x=0$ intersection. Why: Students rush to simplify the equation and forget $x=0$ is a valid solution, leaving them with one fewer intersection than needed. Correct move: Always rearrange to $f(x)-g(x)=0$, factor out all common terms first, then solve for all roots.
• Wrong move: Integrating $f(x)-g(x)$ over the entire interval from leftmost to rightmost intersection, without splitting at intermediate intersections. Why: Students confuse net signed area with total area, so negative areas from swapped curves cancel out positive areas, leading to a much too small result. Correct move: Always split the interval at every intersection, and integrate upper minus lower on each subinterval separately.
• Wrong move: Assuming upper function automatically alternates after each intersection, skipping the test step. Why: Students memorize that polynomials alternate, but this fails for non-polynomials or functions with double roots, leading to wrong upper/lower assignments. Correct move: Always test a point inside each subinterval to confirm which function is upper before setting up the integral.
• Wrong move: Writing (lower function - upper function) for a subinterval, leading to negative area contribution. Why: Students mix up the sign of the difference when testing, leading to a lower total area than the correct value. Correct move: Confirm that your (upper - lower) difference is positive at the test point before integrating.
• Wrong move: Only calculating the area between each consecutive pair of intersections as separate regions, leading to double-counting or including unbounded area. Why: Students misinterpret "total area bounded by the two curves" and misidentify the full interval of integration. Correct move: Always use the leftmost and rightmost intersection as the outer bounds of integration, splitting at all intermediate intersections.

Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

The curves $y=x^3-5x$ and $y=3x$ intersect at $x=-2\sqrt{2}$, $x=0$, and $x=2\sqrt{2}$. Which of the following is the correct integral set-up for the total bounded area between the two curves? A) ${\int }_{-2\sqrt{2}}^{2\sqrt{2}}(x^3-5x-3x)dx$ B) ${\int }_{-2\sqrt{2}}^0(x^3-8x)dx+{\int }_0^{2\sqrt{2}}(8x-x^3)dx$ C) ${\int }_{-2\sqrt{2}}^0(8x-x^3)dx+{\int }_0^{2\sqrt{2}}(x^3-8x)dx$ D) ${\int }_0^{2\sqrt{2}}(x^3-8x{)}^{2}dx$

Worked Solution: First, rewrite the difference $f(x)-g(x)=(x^3-5x)-3x=x^3-8x$. Test the first subinterval $(-2\sqrt{2},0)$ with $x=-2$: $f(-2)-g(-2)=(-8)-8(-2)=8>0$, so $f(x)$ is upper, meaning the integrand for the first interval is $x^3-8x$. Test the second subinterval $(0,2\sqrt{2})$ with $x=2$: $f(2)-g(2)=8-16=-8<0$, so $g(x)$ is upper, meaning the integrand for the second interval is $3x-(x^3-5x)=8x-x^3$. This matches option B. Option A is wrong because it does not split the interval, so it calculates net signed area (which equals 0). Option C swaps upper and lower. Option D uses an incorrect formula for area. Correct answer: $\boxed{B}$

Question 2 (Free Response)

Let $f(x)=x^4-5x^2$ and $g(x)=-4$. (a) Find all $x$-coordinates of the intersections of $f(x)$ and $g(x)$. (b) Identify the upper function on each subinterval between the leftmost and rightmost intersections. (c) Find the total area of all bounded regions between $f(x)$ and $g(x)$.

Worked Solution: (a) Set $f(x)=g(x)$: $x^4-5x^2=-4⟹x^4-5x^2+4=0⟹(x^2-1)(x^2-4)=0$ Solutions: $x^2=1⟹x=\pm 1$, $x^2=4⟹x=\pm 2$. All intersection $x$-coordinates: $-2,-1,1,2$.

(b) Sorted intersections: $-2<-1<1<2$, giving three subintervals:

• $[-2,-1]$: Test $x=-1.5$: $f(-1.5)-g(-1.5)=(5.0625-11.25)+4=-2.1875<0$, so $g(x)$ is upper.
• $[-1,1]$: Test $x=0$: $f(0)-g(0)=0+4=4>0$, so $f(x)$ is upper.
• $[1,2]$: Test $x=1.5$: Same as $x=-1.5$, $f-g=-2.1875<0$, so $g(x)$ is upper.

(c) Set up and evaluate the area integral: $$A={\int }_{-2}^{-1}(-4-(x^4-5x^2))dx+{\int }_{-1}^1(x^4-5x^2+4)dx+{\int }_1^2(-x^4+5x^2-4)dx$$ By symmetry, the first and third integrals are equal: $2\cdot \frac{22}{15}=\frac{44}{15}$. The middle integral evaluates to $\frac{76}{15}$. Total area: $A=\frac{44+76}{15}=8$.

Question 3 (Application / Real-World Style)

A city is developing a park between a river and a proposed walking trail. The river’s path follows $R(t)=t^3-6t^2+11t$, and the trail follows $T(t)=6t$, for $0\le t\le 5$, where both $R(t)$ and $t$ are measured in hundreds of meters. Find the total area of the park (the region between the river and trail between the southernmost and northernmost intersections) and give your answer in square meters.

Worked Solution:

1. Find intersections: Set $t^3-6t^2+11t=6t⟹t^3-6t^2+5t=0⟹t(t-1)(t-5)=0$, so sorted intersections are $t=0,t=1,t=5$.
2. Identify upper/lower: On $[0,1]$, test $t=0.5$: $R-T=1.125>0$, so $R$ is upper. On $[1,5]$, test $t=2$: $R-T=-6<0$, so $T$ is upper.
3. Set up and evaluate: $$A={\int }_0^1(t^3-6t^2+5t)dt+{\int }_1^5(-t^3+6t^2-5t)dt=\frac{3}{4}+31=31.75$$
4. Unit conversion: Area is in units of (hundreds of meters)², so $31.75\times (100{)}^2=317500$ square meters.

The new city park will have a total area of 317,500 square meters.

Quick Reference Cheatsheet

What's Next

This topic is the foundation for more advanced integration applications later in Unit 8, starting with calculating volumes of solids of revolution and volumes with known cross-sections. For these topics, you will need to find the distance between two curves (which depends on correctly identifying upper/lower curves across intervals with multiple intersections) before integrating for volume. Without mastering the process of splitting intervals at multiple intersections and tracking upper/lower curves, you will not be able to correctly set up these volume integrals, which make up a large portion of Unit 8 exam points. This topic also reinforces the key distinction between net signed area and total area, which is critical for all accumulation problems in context.

Volumes of solids of revolution Volumes with known cross-sections Accumulation functions in context Average value of a function